Let's take as example frames the GCRF (inertial, Earth-centered, non rotating) and ITRF (non-inertial, Earth-centered, rotating).

Previous questions have discussed the conversion of position and velocity.

However, there is still one key component missing, which is the conversion of acceleration vectors between the two frames. I believe this is in fact key, since, for example, an accurate acceleration due to the Earth gravitational attraction can be calculated by using spherical harmonics, which will result in an acceleration in ITRF which needs to be converted back to GCRF.

I have checked Vallado's Fundamentals of Astrodynamics and Applications, section 3.4, where the following equation is presented (the original equation uses subindices Inertial and Rot instead of GCRF and ITRF respectively; I have replaced them here by GCRF and ITRF, as they are the examples of inertial and rotating frames that I am currently considering):

$$ \vec{a}_{GCRF} = \vec{a}_{ITRF} + \dot{\vec{\omega}}\ \times\ \vec{r}_{ITRF} + \vec{\omega}\ \times\ \left(\vec{\omega}\ \times\ \vec{r}_{ITRF} \right) + 2\ \vec{\omega}\ \times\ \vec{v}_{ITRF} $$


  • $\vec{a}_{GCRF}$ is the acceleration in GCRF frame
  • $\vec{a}_{ITRF}$ is the acceleration in ITRF frame
  • $\vec{r}_{ITRF}$ is the position in ITRF frame
  • $\vec{v}_{ITRF}$ is the velocity in ITRF frame
  • $\vec{\omega}$ is the angular velocity vector of Earth, which as David Hammen explained in his answer here, can be quite difficult to calculate in a very exact manner, but approximate ways seem relatively straightforward
  • $\dot{\vec{\omega}}$ would be, if I understand correctly, the time-derivative of said angular velocity vector, i.e., a measurement of the rate of change of the rotation of Earth

I thought that $\dot{\vec{\omega}}$ would be very close to 0, and therefore that term would be ignored. However, Vallado goes on after that equation to state:

Notice the rotating acceleration is augmented by several other parameters to account for the coordinate system rotation. The supplemental terms are usually identified with specific names. The first term ($\dot{\vec{\omega}}\ \times\ \vec{r}_{ITRF}$) is called the tangential acceleration because $\vec{\omega}$ changes. For the satellite problems, it’s zero only for a true circular orbit. Next are the centripetal acceleration, ($\vec{\omega}\ \times\ \left(\vec{\omega}\ \times\ \vec{r}_{ITRF} \right)$), and the Coriolis acceleration, ($2\ \vec{\omega}\ \times\ \vec{v}_{ITRF}$)

(Emphasis added by me; I have also modified the exact mathematic notation used in the quote to match the equation with modified notation I presented previously).

It seems then that such term would in fact not be 0 for all orbits that are not circular. But I am surprised by this, why would this term be 0 or different than 0 depending on the eccentricity of the orbit of the satellite? I thought it would be independent of the satellite, since it is a measurement of the rate of change of Earth's angular velocity.

To add to the confusion, I have found other sources, such as this thesis, which state basically that the acceleration can be directly converted between GCRF and ITRF frames by simply applying the same transformation that is applied to position vectors (see equation 2.73 at the end of page 30 of the linked document and the preceding paragraph).

So, which would be the correct procedure? If the procedure presented by Vallado is in fact the correct one (which, in principle, makes more sense to me), what is exactly the term involving $\dot{\vec{\omega}}\ \times\ \vec{r}_{ITRF}$?

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    $\begingroup$ My guess is that "For the satellite problems, it’s zero only for a true circular orbit" applies to situations when the rotating frame is determined by the movement of some satellite. For example, the frame is centered on the satellite, one axis points to the zenith and another one is perpendicular to the orbital plane. And then, let's say, there is another satellite chasing this one, and you want to write its equations of motion in this frame. $\endgroup$
    – Litho
    Commented Feb 3, 2022 at 8:46
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    $\begingroup$ I see, that makes sense! Thanks a lot! $\endgroup$
    – Rafa
    Commented Feb 3, 2022 at 9:03
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    $\begingroup$ Why are you worried about acceleration in a rotating frame, and why are you worried about what most call the fictions Euler force? You might have a bit of an XY problem going on here. If you're trying to build a self-navigating car, you don't need to worry about that term as $\dot\omega$ is very small for the Earth. The effects are orders of magnitude smaller than measurement uncertainties. $\endgroup$ Commented Feb 3, 2022 at 9:18
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    $\begingroup$ If you're trying to propagate a satellite in some ECEF, my advice is the same advice a doctor gives to a patient who comes in and says "Doc! It hurts when I do this!" The patient then balls one of his hands into a fist and hits himself hard on his head. The advice from the doctor is the same advice about propagating in ECEF: Don't do that. $\endgroup$ Commented Feb 3, 2022 at 9:20
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    $\begingroup$ Ignoring frame transformations, acceleration due to gravity is the same in all Newtonian frames. The same applies to atmospheric drag. Both are real forces in Newtonian mechanics, and real forces in Newtonian mechanics are the same vector in all Newtonian frames. It's only the fictitious forces that differ from one Newtonian frame to another. Note that I've used "Newtonian" multiple times; things get a bit more complex if one wants to consider relativistic effects. $\endgroup$ Commented Feb 3, 2022 at 10:04


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